If $f(x)=\frac{x^5-1}3$, find $f^{-1}(-31/96)$.
The value $x=f^{-1}(-31/96)$ is the solution to $f(x)=-31/96$.  This means \[\frac{x^5-1}3=\frac{-31}{96}.\]Multiplying by 3 gives \[x^5-1=\frac{-31}{32}.\]If we add 1 we get  \[x^5=\frac{-31}{32}+\frac{32}{32}=\frac1{32},\]and the only value that solves this equation is  \[x=\boxed{\frac12}.\]